Respuesta :
Please, use "^" to denote exponentiation. Your 2x2 should be written as 2x^2.
Your relationships are then f(x,y)=2x^2+3y^2-4x-2
and x^2 + y^2 ≤ 16
Please note that the 2nd relationship represents a circle of radius 4 whose entire interior has been shaded.
Graph this. Then, rewrite f(x,y) in the form f(x,y) = 2x^2 - 4x + 3y^2 -2. Can you identify the shape of the curve in the xy plane representing this function? You'll need to find the points (x,y) in which the graph of f(x,y) intersects the shaded circle x^2 + y^2 ≤ 16.
Think: what do "extreme values of f"
Your relationships are then f(x,y)=2x^2+3y^2-4x-2
and x^2 + y^2 ≤ 16
Please note that the 2nd relationship represents a circle of radius 4 whose entire interior has been shaded.
Graph this. Then, rewrite f(x,y) in the form f(x,y) = 2x^2 - 4x + 3y^2 -2. Can you identify the shape of the curve in the xy plane representing this function? You'll need to find the points (x,y) in which the graph of f(x,y) intersects the shaded circle x^2 + y^2 ≤ 16.
Think: what do "extreme values of f"
The minimum value of the function [tex]f(x,y)[/tex] is [tex]-4[/tex] at the point [tex](1,0)[/tex] and the maximum value of the function [tex]f(x,y)[/tex] is [tex]50[/tex] at the points [tex](-2, -2\sqrt{3}), (-2, 2\sqrt{3})[/tex].
[tex]f(x,y)=2x^2+3y^2-4x-2, g(x,y)=x^2+y^2-16.[/tex]
[tex]f_x (x,y)=4x+0-4-0=4(x-1)[/tex]
[tex]f_y (x,y)=0+6y-0-0=6y[/tex]
[tex]4(x-1)=0, x=1[/tex]
[tex]6y=0, y=0.[/tex]
So, critical points of [tex]f(x,y)[/tex] is [tex](1,0)[/tex].
Put this in [tex]g(x,y)=x^2+y^2-16[/tex].
[tex]g(1,0)=1^2+0^2-16=-15<0[/tex].
So, critical point is in the interior point of the given region.
[tex]g_x (x,y)=2x+0-0=2x[/tex]
[tex]g_y (x,y)=0+2y-0=2y[/tex]
Let there is a parameter [tex]\lambda[/tex] such that [tex]f_x = \lambda g_x[/tex] and [tex]f_y = \lambda g_y[/tex].
[tex]4(x-1)=\lambda (2x)[/tex] and [tex]6y=\lambda (2y)[/tex]
So, [tex]2(x-1)=x\lambda[/tex] and [tex]\lambda=3[/tex]
So, [tex]2(x-1)=3x[/tex]
So, [tex]-2=x[/tex]
Put [tex]x=-2[/tex] in [tex]x^2+y^2=16[/tex], to determine the value of [tex]y[/tex].
[tex](-2)^2+y^2=16[/tex]
[tex]y^2=12[/tex]
[tex]y=\pm 2\sqrt{3}[/tex]
So, the obtained possible extrema are [tex](-2, -2\sqrt{3}), (-2, 2\sqrt{3})[/tex].
Now, find [tex]f(x,y)[/tex] at [tex](1,0), (-2, -2\sqrt{3}),[/tex] and [tex](-2, 2\sqrt{3})[/tex].
[tex]f(1,0)=2(1)^2+3(0)^2-4(1)-2[/tex]
[tex]=2-4-2[/tex]
[tex]=-4[/tex]
[tex]f(-2,-2\sqrt{3})=2(-2)^2+3(-2\sqrt{3})^2-4(-2)-2[/tex]
[tex]=8+36+8-2[/tex]
[tex]=50[/tex]
[tex]f(-2,2\sqrt{3})=2(-2)^2+3(2\sqrt{3})^2-4(-2)-2[/tex]
[tex]=8+36+8-2[/tex]
[tex]=50[/tex]
So, the minimum value of the function [tex]f(x,y)[/tex] is [tex]-4[/tex] at the point [tex](1,0)[/tex] and the maximum value of the function [tex]f(x,y)[/tex] is [tex]50[/tex] at the points [tex](-2, -2\sqrt{3}), (-2, 2\sqrt{3})[/tex].
Learn more about extrema here:
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